Count number of increasing subsequences of size k

Given an array arr[] containing n integers. The problem is to count number of increasing subsequences in the array of size k.

Examples:

Input : arr[] = {2, 6, 4, 5, 7}, 
            k = 3
Output : 5
The subsequences of size '3' are:
{2, 6, 7}, {2, 4, 5}, {2, 4, 7},
{2, 5, 7} and {4, 5, 7}.

Input : arr[] = {12, 8, 11, 13, 10, 15, 14, 16, 20}, 
            k = 4
Output : 39

Approach: The idea is to use Dynamic Programming by define 2D matrix, say dp[][]. dp[i][j] stores the count of increasing subsequences of size i ending with element arr[j]. So dp[i][j] can be defined as:

dp[i][j] = 1, where i = 1 and 1 <= j <= n
dp[i][j] = sum(dp[i-1][j]), where 1 < i <= k, i <= j <= n and arr[m] < arr[j] for (i-1) <= m < j.



Below is the implementation of above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation to count number of
// increasing subsequences of size k
#include <bits/stdc++.h>
  
using namespace std;
  
// function to count number of increasing
// subsequences of size k
int numOfIncSubseqOfSizeK(int arr[], int n, int k)
{
    int dp[k][n], sum = 0;
    memset(dp, 0, sizeof(dp));
  
    // count of increasing subsequences of size 1
    // ending at each arr[i]
    for (int i = 0; i < n; i++)
        dp[0][i] = 1;
  
    // building up the matrix dp[][]
    // Here 'l' signifies the size of
    // increassing subsequence of size (l+1).
    for (int l = 1; l < k; l++) {
  
        // for each increasing subsequence of size 'l'
        // ending with element arr[i]
        for (int i = l; i < n; i++) {
  
            // count of increasing subsequences of size 'l'
            // ending with element arr[i]
            dp[l][i] = 0;
            for (int j = l - 1; j < i; j++) {
                if (arr[j] < arr[i])
                    dp[l][i] += dp[l - 1][j];
            }
        }
    }
  
    // sum up the count of increasing subsequences of
    // size 'k' ending at each element arr[i]
    for (int i = k - 1; i < n; i++)
        sum += dp[k - 1][i];
  
    // required number of increasing
    // subsequences of size k
    return sum;
}
  
// Driver program to test above
int main()
{
    int arr[] = { 12, 8, 11, 13, 10, 15, 14, 16, 20 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 4;
  
    cout << "Number of Increasing Subsequences of size "
         << k << " = " << numOfIncSubseqOfSizeK(arr, n, k);
  
    return 0;
}

chevron_right






                        filter_none









Java














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















//Java implementation to count number of 


// increasing subsequences of size k 


  


class GFG { 


  


// function to count number of increasing 


// subsequences of size k 


    static int numOfIncSubseqOfSizeK(int arr[], int n, int k) { 


        int dp[][] = new int[k][n], sum = 0; 


  


        // count of increasing subsequences of size 1 


        // ending at each arr[i] 


        for (int i = 0; i < n; i++) { 


            dp[0][i] = 1; 


        } 


  


        // building up the matrix dp[][] 


        // Here 'l' signifies the size of 


        // increassing subsequence of size (l+1). 


        for (int l = 1; l < k; l++) { 


  


            // for each increasing subsequence of size 'l' 


            // ending with element arr[i] 


            for (int i = l; i < n; i++) { 


  


                // count of increasing subsequences of size 'l' 


                // ending with element arr[i] 


                dp[l][i] = 0; 


                for (int j = l - 1; j < i; j++) { 


                    if (arr[j] < arr[i]) { 


                        dp[l][i] += dp[l - 1][j]; 


                    } 


                } 


            } 


        } 


  


        // sum up the count of increasing subsequences of 


        // size 'k' ending at each element arr[i] 


        for (int i = k - 1; i < n; i++) { 


            sum += dp[k - 1][i]; 


        } 


  


        // required number of increasing 


        // subsequences of size k 


        return sum; 


    } 


  


// Driver program to test above 


    public static void main(String[] args) { 


        int arr[] = {12, 8, 11, 13, 10, 15, 14, 16, 20}; 


        int n = arr.length; 


        int k = 4; 


  


        System.out.print("Number of Increasing Subsequences of size "


                + k + " = " + numOfIncSubseqOfSizeK(arr, n, k)); 


  


    } 


} 


// This code is contributed by 29AjayKumar 



















                        chevron_right







                        filter_none









Python3














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















# Python3 implementation to count number  


# of increasing subsequences of size k 


import math as mt 


  


# function to count number of increasing 


# subsequences of size k 


def numOfIncSubseqOfSizeK(arr, n, k): 


  


    dp = [[0 for i in range(n)]  


             for i in range(k)] 


               


    # count of increasing subsequences  


    # of size 1 ending at each arr[i] 


    for i in range(n): 


        dp[0][i] = 1


  


    # building up the matrix dp[][] 


    # Here 'l' signifies the size of 


    # increassing subsequence of size (l+1). 


    for l in range(1, k): 


  


        # for each increasing subsequence of  


        # size 'l' ending with element arr[i] 


        for i in range(l, n): 


  


            # count of increasing subsequences of  


            # size 'l' ending with element arr[i] 


            dp[l][i] = 0


            for j in range(l - 1, i): 


                if (arr[j] < arr[i]): 


                    dp[l][i] += dp[l - 1][j] 


              


    # Sum up the count of increasing subsequences  


    # of size 'k' ending at each element arr[i] 


    Sum = 0


    for i in range(k - 1, n): 


        Sum += dp[k - 1][i] 


  


    # required number of increasing 


    # subsequences of size k 


    return Sum


  


# Driver Code 


arr = [12, 8, 11, 13, 10, 


          15, 14, 16, 20 ] 


n = len(arr) 


k = 4


  


print("Number of Increasing Subsequences of size", 


         k, "=", numOfIncSubseqOfSizeK(arr, n, k)) 


  


# This code is contributed by 


# Mohit kumar 29 



















                        chevron_right







                        filter_none









C#














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















// C# implementation to count number of 


// increasing subsequences of size k 


   


using System; 


                      


public class GFG { 


   


// function to count number of increasing 


// subsequences of size k 


    static int numOfIncSubseqOfSizeK(int []arr, int n, int k) { 


        int [,]dp = new int[k,n]; int sum = 0; 


   


        // count of increasing subsequences of size 1 


        // ending at each arr[i] 


        for (int i = 0; i < n; i++) { 


            dp[0,i] = 1; 


        } 


   


        // building up the matrix dp[,] 


        // Here 'l' signifies the size of 


        // increassing subsequence of size (l+1). 


        for (int l = 1; l < k; l++) { 


   


            // for each increasing subsequence of size 'l' 


            // ending with element arr[i] 


            for (int i = l; i < n; i++) { 


   


                // count of increasing subsequences of size 'l' 


                // ending with element arr[i] 


                dp[l,i] = 0; 


                for (int j = l - 1; j < i; j++) { 


                    if (arr[j] < arr[i]) { 


                        dp[l,i] += dp[l - 1,j]; 


                    } 


                } 


            } 


        } 


   


        // sum up the count of increasing subsequences of 


        // size 'k' ending at each element arr[i] 


        for (int i = k - 1; i < n; i++) { 


            sum += dp[k - 1,i]; 


        } 


   


        // required number of increasing 


        // subsequences of size k 


        return sum; 


    } 


   


// Driver program to test above 


    public static void Main() { 


        int []arr = {12, 8, 11, 13, 10, 15, 14, 16, 20}; 


        int n = arr.Length; 


        int k = 4; 


   


        Console.Write("Number of Increasing Subsequences of size "


                + k + " = " + numOfIncSubseqOfSizeK(arr, n, k)); 


   


    } 


} 


// This code is contributed by 29AjayKumar 



















                        chevron_right







                        filter_none









PHP














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















<?php 


// PHP implementation to count  


// number of increasing  


// subsequences of size k 


  


// function to count number 


// of increasing subsequences 


// of size k 


function numOfIncSubseqOfSizeK($arr,  


                               $n, $k) 


{ 


    $dp = array(array());  


    $sum = 0; 


    $dp = array_fill(0, $n + 1, false); 


  


    // count of increasing  


    // subsequences of size 1 


    // ending at each arr[i] 


    for ($i = 0; $i < $n; $i++) 


        $dp[0][$i] = 1; 


  


    // building up the matrix  


    // dp[][]. Here 'l' signifies  


    // the size of increasing  


    // subsequence of size (l+1). 


    for ($l = 1; $l < $k; $l++) 


    { 


  


        // for each increasing  


        // subsequence of size 'l' 


        // ending with element arr[i] 


        for ($i = $l; $i < $n; $i++) 


        { 


  


            // count of increasing  


            // subsequences of size 'l' 


            // ending with element arr[i] 


            $dp[$l][$i] = 0; 


            for ($j = $l - 1; $j < $i; $j++)  


            { 


                if ($arr[$j] < $arr[$i]) 


                    $dp[$l][$i] += $dp[$l - 1][$j]; 


            } 


        } 


    } 


  


    // sum up the count of increasing  


    // subsequences of size 'k' ending 


    // at each element arr[i] 


    for ($i = $k - 1; $i < $n; $i++) 


        $sum += $dp[$k - 1][$i]; 


  


    // required number of increasing 


    // subsequences of size k 


    return $sum; 


} 


  


// Driver Code 


$arr = array(12, 8, 11, 13,  


             10, 15, 14, 16, 20); 


$n = sizeof($arr); 


$k = 4; 


  


echo "Number of Increasing ".  


     "Subsequences of size ", 


                 $k , " = " ,  


     numOfIncSubseqOfSizeK($arr,  


                           $n, $k); 


  


// This code is contributed  


// by akt_mit 


?> 



















                        chevron_right







                        filter_none










Output:


Number of Increasing Subsequences of size 4 = 39



Time Complexity: O(kn2).

Auxiliary Space: O(kn).


Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.





    

        My Personal Notes
        arrow_drop_up
    

    

        
        

            
            
        

    


Recommended Posts:
Image
ayushjauhari14
Check out this Author's contributed articles.
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on  the "Improve Article" button below.